tailieunhanh - Electromagnetic Waves and Antennas combined - Chapter 2

Uniform Plane Waves in Lossless Media Bởi vì cũng ∂ z Ez = 0, sau đó Ez phải có một hằng số, độc lập của z, t. Không bao gồm các giải pháp tĩnh, chúng tôi có thể thực hiện việc này liên tục bằng không. Tương tự như vậy, chúng tôi có Hz = 0. Do đó, các lĩnh vực có các thành phần dọc theo x, y hướng dẫn: E (z, t) = x Ex (z, t) + y EY (z, t) H (z, t) = x HX (z, t) + y Hy (z, t) (các lĩnh. | 2 Uniform Plane Waves Uniform Plane Waves in Lossless Media The simplest electromagnetic waves are uniform plane waves propagating along some fixed direction say the z-direction in a lossless medium e jU . The assumption of uniformity means that the fields have no dependence on the transverse coordinates x y and are functions only of z f. Thus we look for solutions of Maxwell s equations of the form E x y z f E z t and H x y z t H z t . Because there is no dependence on X y we set the partial derivatives sx 0 and Sy 0. Then the gradient divergence and curl operations take the simplified forms v z . SEZ V - Sz Vx zXtt -x Sz SEy Sz Assuming that D eE and B fjH the source-free Maxwell s equations become .SH St V 0 V H 0 SE SH z X Sz SH SE z X e Sz St SEZ n 0 Sz 0 Sz An immediate consequence of uniformity is that and H do not have components along the z-direction that is Ez Hz 0. Taking the dot-product of Ampere s law with the unit vector z and using the identity z z X A 0 we have The shorthand notation dv stands for E-. ox . Uniform Plane Waves in Lossless Media 37 Because also SZEZ 0 it follows that Ez must be a constant independent of z t. Excluding static solutions we may take this constant to be zero. Similarly we have Elz 0. Thus the fields have components only along the x y directions z f x x z f ỹ y z f H z f xHx z f ỳHy z f transverse fields These fields must satisfy Faraday s and Ampere s laws in Eqs. . We rewrite these equations in a more convenient form by replacing e and jU by e ỊẲ where c Thus c q are the speed of light and characteristic impedance of the propagation medium. Then the first two of Eqs. may be written in the equivalent forms A. Be z X Sz BH I ZX Sz The first may be solved for SZE by crossing it with z. Using the BAC-CAB rule and noting that has no z-component we have f .SEX BE t SE SE zx xz z-z -z z- SzJ Sz SzJ Sz 1 . ÕH 1 ỀĨ. c St where we used z SZE SZEZ 0 and z z 1. It follows that Eqs. .